Question: 2 Repeated Digits

Comment on 2 Repeated Digits

Hello Brent,

I employed a different method to get the solution. Please let me know if correct.

Step 1 : Total number of 3 digits greater than 499 = 500(5*10*10)
Step 2 : Total number of 3 digits with similar number = 5(555,666,777,888,999)
Step 3 : Total number of 3 digits with exactly one number same = 360(5*9*8)
Step 4 : Total number of 3 digits with exactly two numbers same = 500 - 5 - 360 = 135
greenlight-admin's picture

Perfect approach!

Great approach!!
Your method made me more clear about dealing such questions.

Thank you.

Why is 922 not included?
greenlight-admin's picture

922 is just one of the many possible outcomes I listed in order to get a better idea of what the possible outcomes look like.

Once I listed some outcomes, I saw that there are three CATEGORIES of outcomes that satisfy the given requirements.

Those three categories are:
1) different, same, same
2) same, different, same
3) same, same, different

So, the 45 outcomes in category 1 include numbers like 511, 744, 877, etc. This category ALSO includes 922.
So, 922 IS included among the 45 outcomes in category 1.

Does that help?


My only problem with this is why there should be 1 at last stage. 5*9*1. Please, kindly explain that to me. Why no 5*9*8. Thanks
greenlight-admin's picture

In stage 3, the last digit must be the SAME as another number.
For example, at 1:50, the 3rd digit must the same as the 2nd digit.

So, if during stage 2, the digit 7 was selected, then the 3rd digit must be 7.
So, there's only ONE way to select the 3rd digit so that it matches the 2nd digit.

Does that help?


Digit must be greater than 499 i.e starts 500 to 999

Now rules -> _ _ _
possible values for 1st digit {5,6,7,8,9}
And second, third digits has {0,1,2,3,4,5,6,7,8}

Lets start hunting here-
Case 1:
_ _ _ Here R -> Repeat digit
R ? R

Case 2:
_ _ _
? R R

Case 3:
_ _ _
R R ?

In Case 1:
I pick 1 of 5 possible values in first digit -5 ways
Second digit 1 of 10 possible values - 9 ways(because already one number is used in 1st place and that should not repeat again here)
Third digit in 1 way as that should be same as first digit with no exception.

Now 5*9*1 = 45 by counting principle.

case 1: 45...
As I have 2 more such cases

45+45+45 = 135

greenlight-admin's picture


I know this might take longer, but I decided to write out all the numbers between 500 and 599 that satisfies the conditions. I found 27 numbers between 500 and 599 that apply. Therefore, the same amount of numbers exist between 600 and 699, 700 and 799, 800 and 899, and 900 and 999. These are 5 sets of number ranges, so 27 x 5 = 135. Bingo. Yes, this takes longer, but at least you don't have to write out every single number. Once you detect a pattern in one range of numbers, it will apply to the other ranges. A correct answer is a correct answer is a correct answer, no matter which way you slice it. Time is the only enemy in a case like this.
greenlight-admin's picture

Great strategy!
In my opinion, not enough students consider listing and counting as a viable approach. More often than not, a pattern will emerge, and you can use that pattern to arrive at the correct answer.

could you explain how this will work with the restrictions rule? where you ignore restrictions, then subtract the ways to break the restriction?
greenlight-admin's picture

Using the Restriction Rule is tricky here because NOT having 2 repeated digits means two things:
1) having zero repeated digits
2) having 3 repeated digits

We'll have to calculate each one individually

1) having zero repeated digits
The hundreds digit can be 5, 6, 7, 8, or 9. There are 5 options.
The tens digit can be any digit other than the one already selected. There are 9 options.
The units digit can be any digit other than the two already selected. There are 8 options.
Total number of 3-digit integers with of 0 repeated digits = (5)(9)(8) = 360

2) having 3 repeated digits
Let's list them: 555, 666, 777, 888, 999
Total = 5

360 + 5 = 365
So, there are 365 different ways to BREAK the rule about having 2 repeated digits

IGNORE the restriction
If we ignore the restriction, then we are dealing with all 500 3-digit integers greater than 499.

So, the number of 3-digit integers with of ZERO repeated digits = 500 - 365 = 135

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